math project presentation new

THE BALLOT PROBLEM OF MANY CANDIDATES

Advisers :

Mr.Thammanoon Puirod

Mr. Deaw Jaibun

Members :

Mr. Norrathep Rattanavipanon

Mr. Nuttakiat Chaisettakul

Mr. Papoj Thamjaroenporn

What is the Ballot problem?What is the Ballot problem?

Suppose A and B are candidates and there are a+b voters, a voting for A and b for B. In how many ways can the ballots be counted so that A is always ahead of B ?

a + ba - b

aa + b

The formula comes up to be .

a voters for A

b voters for B

Why is it interesting?Why is it interesting?

The Ballot problem is one of very well-known problems. This problem had been proved for two candidates.

What about any n, which is more than 2, candidates?

ObjectiveObjective

To find the formula and proof of Ballot problem for many candidates.

In the case of two candidates

(The Original Ballot problem)(The Original Ballot problem)

Suppose is the ballots of the 1st candidate.

is the ballots of the 2nd candidate, when .

Define “1” as the ballot given to 1st candidate. “-1” as the ballot given to 2nd candidate.

1 2a > a2a1a

In the case of two CandidatesIn the case of two Candidates

The number of ways to count the ballots for

required condition.

The number of permutation of the sequence:

such that the partial sum is always positive.

The number of ways to walk on the lattice plane starting at (0,0) and finish at (a,b), and not allow pass through the line y=x except (0,0)

' '

1,1, ,1, 1, 1, , 1a s b s

=

=

In the case of two CandidatesIn the case of two Candidates

Reflection Principle

1

2

Since the number of ways begin at (0,1) is the permutation of a ' 1s

and (a -1) '(-1)s and vice versa.

“Reflection principle” is use to count the number of path, one can show that the number of illegal ways which begin at (1,0) is equal to the number of ways begin at (0,1). It implies that, if we denote as the number of ways as required:

1 2 1 2 1 21 21 2

1 2 11 2

a + a -1 a + a -1 a + aa - aB(a ,a ) = - =

a -1 a -1 aa + a

1 2B(a ,a )

Reflection Principle

Figure 1

Figure 2

In the case of three Candidates

.

st1

nd2

rd3

1 2 3

a be the ballot of the 1 candidate,

a be the ballot of the 2 candidate,

a be the ballot of the 3 candidate,

when a > a > a

Suppose

In the case of three Candidates

In the case of three Candidates

Figure 3

Figure 4

How to count ?

How to count ?How to count ?

1. Consider the path seen at the front view. Find all possible ways to “walk” within the allowed plane.

2. Consider the path seen at the side view. Find all possible ways to “walk” within the allowed plane.

3. Match each way of 1. to each of 2. (Sometimes it yields more than one complete path.)

How to count ?How to count ?

Example

Counting 1 2 3(a ,a ,a ) = (4,3,2)

Front ViewFront View

F(1) F(2) F(3)

F(4) F(5)

Example

Counting 1 2 3(a ,a ,a ) = (4,3,2)

Side ViewSide View

S(1) S(2)

Example

Counting 1 2 3(a ,a ,a ) = (4,3,2)

MatchingMatching

22111S(2)

11111S(1)

F(5)F(4)F(3)F(2)F(1)F(i) * S(j)

F(1) F(2) F(3)

F(4) F(5)S(1) S(2)

5

7

12

Total

Dynamic ProgrammingDynamic Programming

Counting (5,4) with D.P.

111111

432100

952000

1450000

1400000

y

x

x+y

1

Formula for three candidates

1 21 21 2

11 2

a + aa - aB(a ,a ,0) =

aa + a

Definition is the number of ways to count the ballot that correspond to the required condition

1 2 3B(a ,a ,a )

Lemma 1.1

1 1 2B(a ,a ,a ) = 0Lemma 1.2

Formula for three candidatesFormula for three candidates

1 2 31 3 2 31 21 2 3

1 2 31 2 1 3 2 3

a + a + aa - a a - aa - aB(a ,a ,a ) =

a ,a ,aa + a a + a a + a

Conjecture

Formula for three candidatesFormula for three candidates

1 2 3 1 2 3a ,a ,a N and (a > a > a ) Let

1 2 3 1 2 3 1 2 3 1 2 3B(a ,a ,a ) = B(a -1,a ,a ) + B(a ,a -1,a ) + B(a ,a ,a -1)Consider

Use strong induction; given is the “base”

therefore

B(3,2,1)

Proof

B(3,2,1) = B(2,2,1) + B(3,1,1) + B(3,2,0)

3+ 23- 2 1 5!= 0 + 0 + = ×

23+ 2 5 3!2!

3+ 2 +13-1 2 -1 3- 2= 2 =

3,2,13+1 2 +1 3+ 2

Hence, the base case is true.

Formula for three candidatesFormula for three candidates

We assume that for some

all of the following are true;

1 1

2 2

3 3 1 2 3 1 2 3

0 < k a

0 < k a

0 < k a and (a ,a ,a ) (k ,k ,k ),

1 2 3k ,k andk ,

.1 2 31 3 2 31 21 2 3

1 2 31 2 1 3 2 3

k + k + kk - k k - kk - kB(k ,k ,k ) =

k ,k ,kk + k k + k k + k

We will use this assumption to prove that is true1 2 3B(a ,a ,a )

Formula for three candidatesFormula for three candidates

1 2 3 1 2 3 1 2 3 1 2 3

1 2 31 3 2 31 2

1 2 31 2 1 3 2 3

1 21 3 2 31 2

1 2 1 3 2 3

B(a ,a ,a ) = B(a -1,a ,a ) + B(a ,a -1,a ) + B(a ,a ,a -1)

a -1+ a + aa -1- a a - aa -1- a=

a -1,a ,aa -1+ a a -1+ a a + a

a + a -1a - a a -1- aa - a +1+

a + a -1 a + a a -1+ a

3

1 2 3

1 2 31 3 2 31 2

1 2 31 2 1 3 2 3

1 2 3 1 2 1 3 2 3 1

1 2 3 1 2 1 3 2 3 1 2 3

1 2

+ a

a ,a -1,a

a + a + a -1a - a +1 a - a +1a - a+

a ,a ,a -1a + a a + a -1 a + a -1

a + a + a (a -1- a )(a -1- a )(a - a )a= +

a ,a ,a (a -1+ a )(a -1+ a )(a + a )(a + a + a )

(a +1- a )

1 3 2 3 2 1 2 1 3 2 3 3

1 2 1 3 2 3 1 2 3 1 2 1 3 2 3 1 2 3

1 2 3 2 3 1 31 2

1 2 3 1 2 2 3 1 3

(a - a )(a -1- a )a (a - a )(a +1- a )(a - a +1)a+

(a -1+ a )(a + a )(a + a -1)(a + a + a ) (a + a )(a -1+ a )(a + a -1)(a + a + a )

a + a + a a - a a - aa - a=

a ,a ,a a + a a + a a + a

Formula for three candidatesFormula for three candidates

1 2 3 2 3 1 31 21 2 3

1 2 3 1 2 2 3 1 3

a + a + a a - a a - aa - aB(a ,a ,a ) = .

a ,a ,a a + a a + a a + a

By strong induction, we get that,

Formula for three candidatesFormula for three candidates

Formula for n candidates

1 2 3 n 1 2 3 nB(a ,a ,a , , a ,0) = B(a ,a ,a , ,a )

is the number of ways to count the ballots of the n candidates such that, while the ballots are being counted, the winner will always get more ballots than the loser.

Definition

Lemma 3

1 2 3 nB(a ,a ,a , ,a )

Formula for n candidatesFormula for n candidates

(base) n=2

n=k

P(n) is true.

1 2 k

i i 1 2 k 1 2 k

Suppose the formula corresponds to B(b ,b ,...,b ),

b a i {1,2,...,k},(b ,b ,...,b ) (a ,a ,...,a )

Mathematical Induction

n

i i ji 11 2 n i i j

0<i< j n i j1 2 n

a a - aP(n) = "B(a ,a ,…,a ) = , n {1},a ,a a ,i < j"

a + aa ,a ,…,a

Sub - Strong Mathematical Induction

1 2 kQ(k) = "the formula corresponds to B(a ,a ,...,a )for n = k"

k-1

i i ji 11 2 k-1

0<i< j k-1 i j1 2 k-1

a a - aB(a ,a ,…,a ,0) =

a + aa ,a ,…,a

(base)

k

1 2 k 1 i ki=1

B(a ,a ,...,a ) = B(a ,...,a -1,...,a )

1 2 kB(a ,a ,...,a ) is true.

n=k-1

Proof

1 2 3 nLet P(n) is "B(a ,a ,a , ,a ) is true" n

In base case P(2) is true.

Assume P(k -1) is true. We will prove that P(k) is true

1 2 3 k i j1 2 3 k

0<i< j k1 2 3 k i j

a +a + a +…+ a a - aB(a ,a ,a ,…,a ) = .

a ,a ,a ,…,a a + a

Formula for n candidatesFormula for n candidates

1 2 3 k i i

1 2 3 k 1 2 3 k

1 2 3 k i j1 2 3 k

0<i< j k1 2 3 k i j

We assume that for b ,b ,b ,…,b , 0 < b a for i = 1,2,...,k

and (b ,b ,b ,…,b ) (a ,a ,a ,…,a ),

all of the following are true :

b b + b +…+ b b bB(b ,b ,b ,…,b ) = .

b ,b ,b ,…,b b + b

1 2 k 1 2 k

1 2 k

1 2 k-1

Let Q(a ,a , ,a ) is "B(a ,a , ,a ) is true" (only for k)

We need to proof that Q(a ,a , ,a ) is true.

In base case, Q(a ,a , ,a ,0) is true.

Formula for n candidatesFormula for n candidates

1 2 k 1 2 k 1 2 k

1 2 k

i m

i mi j

0<i< j n 0<i n m ii ji, j m i m

m i

Induction

Consider B(a ,a ,…,a ) = B(a -1,a ,…,a ) + B(a ,a -1,…,a ) +

B(a ,a ,…,a -1)

a - (a -1),i < m

a + (a -1)a - a =

(a -1) - aa + a,i > m

(a -1) + a

n1 2 n

m=1 1 2 m n

a + a +…+ a

a ,a ,…,a -1,…,a

Formula for n candidatesFormula for n candidates

1 2 ni j

1 2 n0<i< j n i j

i m

1 2 ni j i m

1 2 mm i0<i< j n 0<i ni ji, j m i m

m i

a +a +…+aa -a=

a ,a ,…,aa +a

a -(a -1),i < m

a +a +…+aa -a a +(a -1)

a ,a ,…,a -1,…,a(a -1)-aa +a ,i > m(a -1)+a

n

nm=1

By mathematical induction,

1 2 3 k i j1 2 3 k

0<i< j k1 2 3 k i j

a +a + a +…+ a a - aB(a ,a ,a ,…,a ) = .

a ,a ,a ,…,a a + a

Formula for n candidatesFormula for n candidates

1. Think about the condition “never less than” instead of “always more than.”

DevelopmentDevelopment

2. What if we suppose that the ballots of the 3rd candidate don’t relate to anyone else?

Application In BiologyApplication In Biology

Lead to finding the number of random graph models for angiogenesis in the renal glomerulus.

Steps to form a vascular network1. Budding2. Spliting3. Connecting

n(Budding) n(Connecting)

Application In CryptographyApplication In Cryptography

Define the plaintext (code) used to send the data

Increases the security of the system.

Given , when look left to right,

n(A) n(B) n(C) ...

p- sequence

Reference

Chen Chuan-Chong and Koh Khee-Meng, Principles and Techniques in Combinatorics, World Scientific, 3rd ed., 1999.

Marc Renault, Four Proofs of the Ballot Theorem, U.S.A., 2007.

Michael L. GARGANO, Lorraine L. LURIE Louis V. QUINTAS, and Eric M. WAHL, The Ballot Problem, U.S.A.,2005.

Miklos Bona, Unimodality, Introduction to Enumerative Combinatorics, McGrawHill, 2007.

Sriram V. Pemmaraju, Steven S. Skienay, A System for Exploring Combinatorics and Graph Theory in Mathematica, U.S.A., 2004.